empty set
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pka
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If one knows the rules of formal logic this is a quite easy proof.
The statement that [imath]A\subseteq B[/imath] means [imath](\forall x)[x\in A \Rightarrow x\in B][/imath].
From login we know that a false statement implies any statement.
So we have If [imath]x\in\emptyset[/imath] (a false statemet) [imath]\Rightarrow (\forall A)[x\in A][/imath] is a true statement.
Thus the emptyset is a subset of all sets.
[imath][/imath][imath][/imath][imath][/imath]
Dr.Peterson
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This requires a careful definition of "subset". How has it been defined?
And what have you tried? You know how we're supposed to work here.
blamocur
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Implication is a boolean operation, and there is a truth table for it -- do you know what it is?
Dr.Peterson
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There is a lot that can be discussed about this question, and I would like to do so.
It will be easier if you state your opinion in full so we can talk about the details that matter to you. It appears that you believe we should not define "subset" in such a way that the empty set is a subset of every set (though you haven't yet stated the definition you are using, which is important in this discussion). Am I right about that? What reasons can you give for your opinion?
pka
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You simply do not know much about this subject.
If you want to disagree that is fine.
But know that I had had forty five+years of teaching university level(graduate & undergraduate) symbolic logic.
blamocur
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To paraphrase an American politician: you are entitled to your own opinion, but you are not entitled to your own Math.
NO you do not know the laws of symbolic logic other wise you should know that there is no a law of logic in propositional calculus or in predicate calculus that says that:
"A false statement implies any statement"
Refer me to a symbolic logic book that says so
Unless you mean : q&(~q) implies any statement
But that is contradiction implies any statatement
You CANNOT say that : PARIS is in England logicaly implies that 3+4=5
You can say that: Paris is in England implies that 3+4=5
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No implication [math]\implies[/math] is a symbol of the propositional language ,like v,^
Boolean algebra is just another kind of algebra that uses the symbolic logic in the development of its theorems
blamocur
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Truth table - Wikipedia
Dr.Peterson
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Vacuous truth - Wikipedia
Empty Set is Subset of All Sets - ProofWiki
our main topic here is whether the phrase : "a false statement implies any statement" is a law of logic or just comes from the truth table concerning simple implication
"false implie truth" is true and also : "false implies false" is true
Because most of marhematitians have no idea about the laws of logic and how are they applied in each and every proof they invended the Vacuosly true solutionVacuous truth - Wikipedia
en.wikipedia.org
Empty Set is Subset of All Sets - ProofWiki
proofwiki.org
Then to show clearly that we have to go a litle deeper and define what is a mathematical proof
By your title i suppose that you teach students do you teach them thru wiki ?
I will answer tomorrowTruth table - Wikipedia
so the phrase a false statement implies any statement. is correct?To paraphrase an American politician: you are entitled to your own opinion, but you are not entitled to your own Math.
then according to that we can say :
since 2<1 is false this can imply humans are monkies
Are we all monkies here
HERE is the main difference between a proof and a mere implication
2<1 implies humans are monkies is true
But this is not a proof
Without at all agreeing as to tone or personal attacks, I, like chrislav, find this a singularly unpersuasive proof.
Obviously, I do not disagree with the following definition
[math]A \subset B \iff \forall(x)[x \in A \implies x \in B].[/math]
For reasons that I am not sure I find compelling, I understand that logicians have found it useful to define the proposition “if all pigs are positive integers, then pi is a rational number” as true. More generally, they have defined
[math]\{p \implies q \} \text { is true unless } p \text { is true and } q \text { is false.}[/math]
That in turn entails that [math]p \text { is false} \implies \{ p \implies q\} \text { is true.}[/math]
Whether the reasons are compelling or not, I can accept this definition, no matter how apparently absurd the result, because it is harmless.
[math]\{p \implies q \land \neg p\} \not \implies q.[/math]
In short, although the proposition “2 is a member of the empty set entails that for all sets, 2 is a member” is “true“ by definition, but it does not immediately or obviously follow that the proposition “the empty set is a subset of every set” is true. If this is what chrislav means by saying that the proof given in post 2 is “semantic,” I would rather say that proof is missing something.
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I am trying to understand, not argue. I have neither text. Perhaps if you could send me a picture of the Halmos page via PM, I would understand your proof. (It's fair use.)
You say
Now I could restate this in English by choosing 2 as an instance of x.
(1) If 2 is an element of the empty set, then 2 is an element of every set.
(2) Therefore, the empty set is a subset of every set
I agree that statement 1 is true. What I do not see as obvious is how statement 2 follows from statement 1. It is not even the same consequent, and it is certainly not based on modus ponens.
If I put what I interpret that you are saying into modus ponens form,
[math]p \implies q\\ \neg p\\ \therefore q\\ \therefore r.[/math]
That is not a valid argument. Something is missing.
Now after looking at stackexchange, I have come up with this. What I think is missing is an undisclosed generalization step and a specification of how mathematical implication is defined. Let's start with the latter.
[math]\{p \implies q\} \iff \{\neg (p \land \neg q)\} \iff \{(\neg p) \lor q\}.[/math]
That definition of implication justifies the truth of the proposition that a falsity validly implies anything. (It of course does not fit the common meaning of "implication" of English.)
Now in this case we have a definition that depends on implication.
[math]\mathbb A \text { is a subset of } \mathbb B \iff \{x \in \mathbb A \implies x \in \mathbb B\}.[/math]
So we can rewrite the definition as
[math]\mathbb A \text { is a subset of } \mathbb B \iff \{\neg (x \in \mathbb A) \lor (x \in \mathbb B)\}.[/math]
[math]\text {Let } \mathbb B \text { be an ARBITRARY set.}[/math]
[math]\{ \neg (x \in \emptyset) \lor (x \in \mathbb B)\}.[/math]
That statement is necessarily true whether or not x is an element of set B. The term x "cancels out" as it were. The x is definitely not in the empty set.
[math]\therefore \text {BY DEFINITION, } \emptyset \subset \mathbb B.[/math]
But B was an arbitrary set. Therefore the empty set is a subset of any set.
I think that proof is valid. And I think that is what pka's proof meant. But whether that proof is simple in terms of being obvious is a judgment about the audience.
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